Moment of Inertia

⚛️ Angular Kinematics

Part 1 of 7 — Angular Kinematics

Rotational analogs of linear kinematics:

• $\theta \leftrightarrow x$, $\omega \leftrightarrow v$, $\alpha \leftrightarrow a$

$\omega = \frac{d\theta}{dt}, \quad \alpha = \frac{d\omega}{dt}$

For constant $\alpha$:

• $\omega = \omega_0 + \alpha t$
• $\theta = \omega_0 t + \frac{1}{2}\alpha t^2$
• $\omega^2 = \omega_0^2 + 2\alpha\theta$

Worked Example

A wheel accelerates from rest at $\alpha = 4$ $rad/s^{2}$. Find $\omega$ at $t = 3$ s.

$\omega = \omega_0 + \alpha t = 0 + 4(3) = 12$ rad/s ✅

Concept Check 🎯

Angular Kinematics 🧮

1. A wheel starts from rest with $\alpha = 4$ $rad/s^{2}$. Angular velocity (rad/s) at $t = 3$ s?

2. Same wheel: angular displacement (rad) in 3 s?

3. A wheel has $\omega_0 = 2$ rad/s, $\alpha = 4$ $rad/s^{2}$. Find $\omega$ (rad/s) at $t = 2$ s.

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⚛️ Moment of Inertia

Part 2 of 7 — Moment of Inertia

$I = \sum m_i r_i^2 = \int r^2\,dm$

Common moments of inertia:

• Point mass: $I = mr^2$
• Solid cylinder/disk: $I = \frac{1}{2}mr^2$
• Solid sphere: $I = \frac{2}{5}mr^2$
• Thin rod (center): $I = \frac{1}{12}mL^2$
• Thin hoop: $I = mr^2$

Parallel axis theorem: $I = I_{cm} + md^2$

Worked Example

Find I for a 3 kg point mass at 2 m from the axis.

$I = mr^2 = 3(4) = 12$ $kg\cdot m^{2}$ ✅

Concept Check 🎯

Moment of Inertia 🧮

1. A 3 kg point mass is 2 m from the axis. $I$ $(kg\cdot m^{2})$?

2. A solid disk has $m = 4$ kg, $r = 1$ m. $I$ $(kg\cdot m^{2})$?

3. A solid sphere has $m = 5$ kg, $r = 2$ m. $I$ $(kg\cdot m^{2})$?

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⚛️ Torque

Part 3 of 7 — Torque

$\tau = rF\sin\theta = r_{\perp}F$

$\sum \tau = I\alpha$

This is the rotational analog of $F = ma$.

Torque is positive for counterclockwise rotation and negative for clockwise.

Worked Example

A 10 N force is applied at the end of a 0.5 m wrench perpendicular to it. Find the torque.

$\tau = rF\sin 90° = 0.5(10)(1) = 5$ N·m ✅

Concept Check 🎯

Torque 🧮

1. A 10 N force acts at the end of a 0.5 m wrench (perpendicular). Torque (N·m)?

2. A torque of 12 N·m acts on an object with $I = 2$ $kg\cdot m^{2}$. Angular acceleration $(rad/s^{2})$?

3. Two forces create torques of +7 N·m and -3 N·m. Net torque (N·m)?

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📈 Angular Momentum

Part 4 of 7 — Angular Momentum

Two Forms of Angular Momentum

Here $r_{\perp}$ is the perpendicular distance from the point to the particle's line of motion. A particle moving in a straight line still has angular momentum about any point not on that line.

Torque Changes Angular Momentum

$\tau_{\text{net}} = \frac{dL}{dt} \qquad \Longrightarrow \qquad \tau\,\Delta t = \Delta L \text{ (constant torque)}$

The product $\tau\,\Delta t$ is the angular impulse.

Conservation

If the net external torque on a system is zero, its total angular momentum stays constant:

$I_i\omega_i = I_f\omega_f$

A system that changes shape (a skater pulling in her arms) changes $I$, so $\omega$ changes to keep $L$ fixed. Rotational kinetic energy $K = \frac{L^2}{2I}$ is not conserved in that process.

Worked Example — Clay Ball Sticks to a Rod

A thin uniform rod ($M = 1.2$ kg, length $0.80$ m) can turn freely about a vertical axle through its center and starts at rest. A $0.10$ kg ball of clay moving at $6.0$ m/s, perpendicular to the rod, hits one end and sticks. Find the angular speed just after the collision.

Step 1 — Choose the axle as the reference point. The axle's force has zero moment arm about the axle, so angular momentum about the axle is conserved during the collision (linear momentum is not, because the axle pushes on the rod).

Step 2 — Angular momentum before. Only the clay moves; it travels perpendicular to the rod at $r = 0.40$ m from the axle:

$L_i = mvr = (0.10)(6.0)(0.40) = 0.24$ $kg\cdot m^{2}/s$

Step 3 — Rotational inertia after.

$I_f = \frac{1}{12}ML^2 + mr^2 = \frac{1}{12}(1.2)(0.80)^2 + (0.10)(0.40)^2 = 0.064 + 0.016 = 0.080$ $kg\cdot m^{2}$

Step 4 — Solve.

$\omega_f = \frac{L_i}{I_f} = \frac{0.24}{0.080} = 3.0$ rad/s ✅

Concept Check 🎯

Angular Momentum Practice 🧮

1. A 2.0 kg particle moves at 3.0 m/s along a straight line that passes 0.50 m from point O. What is its angular momentum about O $(kg\cdot m^{2}/s)$?

2. A disk with $I = 0.50$ $kg\cdot m^{2}$ spins at 8.0 rad/s. What is its angular momentum $(kg\cdot m^{2}/s)$?

3. A constant torque of 2.0 N·m acts on that disk for 3.0 s in the direction of its spin. What is its new angular momentum $(kg\cdot m^{2}/s)$?

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⚛️ Rolling Motion

Part 5 of 7 — Rolling Motion

For rolling without slipping: $v_{cm} = R\omega$

Total kinetic energy: $KE_{total} = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I\omega^2$

For a rolling object down an incline: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

Worked Example

A solid sphere ($I = \frac{2}{5}mr^2$) rolls down a 5 m high ramp. Find its speed at the bottom. ($g = 10$ m/s²)

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(v/r)^2 = \frac{7}{10}mv^2$

$v = \sqrt{\frac{10gh}{7}} = \sqrt{\frac{10(10)(5)}{7}} \approx 8.45$ m/s ✅

Concept Check 🎯

Rolling Motion 🧮

1. A wheel rolls at $\omega = 3$ rad/s with $R = 2$ m. What is $v_{cm}$ (m/s)?

2. A ball rolls at $v = 6$ m/s with $R = 2$ m. What is $\omega$ (rad/s)?

3. A solid cylinder ($I = \frac{1}{2}mr^2$) rolls without slipping. If $v_{cm} = 4$ m/s and $m = 2$ kg, what is its translational KE (J)?

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⚛️ Problem-Solving Workshop

Part 6 of 7 — Problem-Solving Workshop

Rotation Problem Strategy

1. Identify the axis of rotation
2. Find the moment of inertia about that axis
3. Calculate net torque
4. Apply $\tau = I\alpha$ or conservation of $L$
5. Connect rotational and translational quantities if rolling

Worked Example

A 20 N force acts tangentially on a disk ($m = 4$ kg, $r = 0.5$ m). Find $\alpha$.

$I = \frac{1}{2}mr^2 = \frac{1}{2}(4)(0.25) = 0.5$ $kg\cdot m^{2}$

$\tau = Fr = 20(0.5) = 10$ N·m

$\alpha = \tau / I = 10 / 0.5 = 20$ $rad/s^{2}$ ✅

Concept Check 🎯

Problem-Solving Workshop 🧮

1. A 20 N tangential force on a disk ($I = 0.5$ $kg\cdot m^{2}$, $r = 0.5$ m). $\alpha$ $(rad/s^{2})$?

2. A torque of 10 N·m acts on a wheel ($I = 2$ $kg\cdot m^{2}$). $\alpha$ $(rad/s^{2})$?

3. A wheel accelerates from rest at 4 $rad/s^{2}$ for 3 s. It then decelerates at 6 $rad/s^{2}$. Time to stop (s)?

Concept Check 🔍

Practice

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⚛️ Review & Applications

Part 7 of 7 — Review & Applications

Key Formulas

• Angular kinematics: $\omega = \omega_0 + \alpha t$
• Moment of inertia: $I = \sum mr^2$
• Torque: $\tau = rF\sin\theta = I\alpha$
• Angular momentum: $L = I\omega$
• Rolling: $v = R\omega$, $KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

Worked Example

A merry-go-round ($I = 500$ $kg\cdot m^{2}$) spins at 2 rad/s. A 50 kg child jumps on at $r = 2$ m. Find the new $\omega$.

$I_f = 500 + 50(4) = 700$ $kg\cdot m^{2}$

$\omega_f = \frac{500(2)}{700} = \frac{1000}{700} \approx 1.43$ rad/s ✅

Concept Check 🎯

Review & Applications 🧮

1. A disk ($I = \frac{1}{2}mr^2$, $m = 4$ kg, $r = 1$ m). What is $I$ $(kg\cdot m^{2})$?

2. $\alpha = \tau / I = 10 / 2 = ?$ $rad/s^{2}$

3. A wheel at 10 rad/s has zero net torque. What is $\omega$ (rad/s) after 100 s?

Concept Check 🔍

Practice

Challenge Question 📋